Q 12-11-101JEE MainJEE Main 2023 (1 Feb, Shift 2)Easy
The threshold frequency of a metal is $f_0$. When the light of frequency $2f_0$ is incident on the metal plate, the maximum velocity of photoelectron is $v_1$. When the frequency of incident radiation is increased to $5f_0$, the maximum velocity of photoelectrons emitted is $v_2$. The ratio of $v_1$ to $v_2$ is
Answer: (A) $\dfrac{v_1}{v_2}=\dfrac12$
$\tfrac12mv^2=h(f-f_0)$: $v_1^2\propto f_0$, $v_2^2\propto4f_0$, so $\dfrac{v_1}{v_2}=\dfrac12$.
Solution by Sreeraj P, M.Sc Physics